Question

How do you find the equation of a line tangent to the function `y=(x^3+3x^2)/x` at x=1?

Answer 1

`y=5x-1`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = 1"`

`y=(x^3+3x^2)/x=x^2+3x`

`rArrdy/dx=2x+3`

`"at x = 1"`

`dy/dx=2+3=5`

`y=4rArr(1,4)" is tangent point"`

`"equation of line is"`

`y-4=5(x-1)`

`rArry=5x-1`

Answer 2

We fill first calculate the value of the derivative at 1.

Explanation:

Let's find the derivative first using the quotient rule :

`dy/dx=((3x^2+6x)x-x^3-3x^2)/x^2`

so at `x=1` we get

`dy/dx|_1=((3+6)-1-3)/1=5`

Let's find `y(1)=4`

Now we can find the equation of the tangent line using the formula

`y-y(1)=y'(1)(x-1)=>y-4=5(x-1)=>y=5x-1`